# In a acute angled triangle, we have $\tan(A)\cdot\tan(B)\cdot\tan(C) \geq 3\sqrt{3}$

How to show:

In a acute angled $\triangle \ ABC$ show that $$\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$$

Any ideas?

• Use $\tan A \cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C$, and the AM-GM inequality. Aug 13, 2011 at 6:45
• The question math.stackexchange.com/questions/8732/… includes the identity Srivatsan mentions. Aug 13, 2011 at 6:56

These may be useful:

• Since $\triangle ABC$ is acute, we have $\tan(A),\tan(B),\tan(C)$ positive.

• By A.M-G.M you have $$\displaystyle \frac{\tan(A)+\tan(B)+\tan(C)}{3} \geq \sqrt[3]{\tan(A)\cdot\tan(B)\cdot\tan(C)}$$

• $\text{The equality holds if the triangle is}$ $\textbf{equilateral.}$

$$A+B=\pi-C$$

\begin{align*} &\tan (A+B)= \tan (\pi-C)\\ &(\tan A+ \tan B)/(1-\tan A \tan B)= (\tan \pi- \tan C)/(1+\tan \pi \tan C)=-\tan C\\ &(\tan A+ \tan B)= -\tan C(1-\tan A \tan B)\\ &\tan A + \tan B= -\tan C+ \tan A \tan B \tan C\\ &\tan A + \tan B+ \tan C= \tan A \tan B \tan C \end{align*}

• Now u can use the am-gm inequality as others have stated. and no problem in using that inequality as all the variables involved iwll be positive quantities in case of an acute angled traingle Aug 13, 2011 at 6:52
• You mean $A+B=\pi - C$, I guess. It would also be good if you indicated where exactly you need to use that $A,B,C \lt \pi/2$.
– t.b.
Aug 13, 2011 at 6:54
• $A+B+C=\pi$, not $2\pi$. Fortunately, $\tan(\pi)=\tan(2\pi)$. Aug 13, 2011 at 6:55

HINT:

Use the AM-GM inequality.